let-u-n-k-1-n-1-k-k-and-H-n-k-1-n-1-k-1-calculate-u-n-interms-of-H-n-2-study-the-convergence-of-u-n-3-study-theconvergence-of-u-n-

Question Number 45792 by maxmathsup by imad last updated on 16/Oct/18

l e t u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{[k]}}{k} a n d H_{n} = \sum_{k = 1}^{n} \frac{1}{k}

1) c a l c u l a t e u_{n} i n t e r m s o f H_{n}

2) s t u d y t h e c o n v e r g e n c e o f (u_{n})

3) s t u d y t h e c o n v e r g e n c e o f Σ u_{n .}

Commented by maxmathsup by imad last updated on 17/Oct/18

3) {l i m}_{n \to + \infty} u_{n} \neq 0 \Rightarrow Σ u_{n} d i v e r g e s . .

Commented by maxmathsup by imad last updated on 17/Oct/18

2) c o n v e r g e n c e o f (u_{n}) ?

w e h a v e u_{2 n} = \frac{1}{2} H_{n} + \frac{1}{2} H_{n - 1} - H_{2 n - 1}

= \frac{1}{2} (l n (n) + γ + o (\frac{1}{n})) + \frac{1}{2} (l n (n - 1) + γ + o (\frac{1}{n})) - l n (2 n - 1) - γ - o (\frac{1}{n})

= l n \sqrt{n (n + 1)} - l n (2 n - 1) + o (\frac{1}{n}) = l n (\frac{\sqrt{n^{2} + n}}{2 n - 1}) + o (\frac{1}{n}) \to - l n (2) (n \to + \infty)

u_{2 n + 1} = \frac{1}{2} H_{n} + \frac{1}{2} H_{n} - H_{2 n + 1} = H_{n} - H_{2 n + 1}

= l n (n) + γ - l n (2 n + 1) - γ + o (\frac{1}{n}) = l n (\frac{n}{2 n + 1}) + o (\frac{1}{n}) \to - l n (2) (n \to + \infty)

s o (u_{n}) c o n v e r g e s a n d {l i m}_{n \to + \infty} u_{n} = - l n (2) .

Commented by maxmathsup by imad last updated on 17/Oct/18

w e h a v e u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{[k]}}{k} \Rightarrow u_{n} = \sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{2 p} - \sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{2 p + 1} b u t

\sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{2 p} = \frac{1}{2} H_{[\frac{n}{2}]}

\sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{2 p + 1} = 1 + \frac{1}{3} + \frac{1}{5} + \dots . + \frac{1}{2 [\frac{n - 1}{2}] + 1}

= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots . . + \frac{1}{2 [\frac{n - 1}{2}]} + \frac{1}{2 [\frac{n - 1}{2}] + 1} - \frac{1}{2} (1 + \frac{1}{2} + \dots . + \frac{1}{2 [\frac{n - 1}{2}]})

= H_{2 [\frac{n - 1}{2}] + 1} - \frac{1}{2} H_{[\frac{n - 1}{2}]} \Rightarrow

u_{n} = \frac{1}{2} H_{[\frac{n}{2}]} + \frac{1}{2} H_{[\frac{n - 1}{2}]} - H_{2 [\frac{n - 1}{2}] + 1}

Answered by arcana last updated on 17/Oct/18

1) c o m o u_{n} = \underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k}}{k}

s i n = 2 t e s p a r (p a r a a l g u n t e n t e r o p o s i t i v o)

u_{n} = \underset{k = 1}{\sum^{n / 2}} \frac{{(- 1)}^{2 k - 1}}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{{(- 1)}^{2 k}}{2 k}

= - \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k}

= - \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \frac{1}{2} H_{n / 2}

t o m a n d o m = 2 k - 1, l u e g o

= - \underset{m = 1}{\sum^{n - 1}} \frac{1}{m} + \frac{1}{2} H_{n / 2} = \frac{1}{2} H_{n / 2} - H_{n - 1}

s i n = 2 t - 1 e s i m p a r, t e n e m o s

u_{n} = \underset{k = 1}{\sum^{(n + 1) / 2}} \frac{{(- 1)}^{2 k - 1}}{2 k - 1} + \underset{k = 1}{\sum^{(n - 1) / 2}} \frac{{(- 1)}^{2 k}}{2 k}

= - \underset{k = 1}{\sum^{(n + 1) / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{(n - 1) / 2}} \frac{1}{2 k}

= - \underset{k = 1}{\sum^{(n + 1) / 2}} \frac{1}{2 k - 1} + \frac{1}{2} H_{(n - 1) / 2}

t o m a n d o m = 2 k - 1

= - \underset{m = 1}{\sum^{n}} \frac{1}{m} + \frac{1}{2} H_{(n - 1) / 2} = - H_{n} + \frac{1}{2} H_{(n - 1) / 2}

Commented by maxmathsup by imad last updated on 17/Oct/18

t h a n k y o u A r k a n a f o r t h i s w o r k \dots

Answered by arcana last updated on 17/Oct/18

3 . c o m o \frac{1}{1 + x} = \underset{k = 0}{\sum^{\infty}} {(- x)}^{k} p a r a ∣ x ∣< 1

i n t e g r a n d o a a m b o s l a d o s v e m o s q u e

\Rightarrow l n (1 + x) = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k + 1} \cdot x^{k + 1}

c o n - 1 < x ⩽ 1 . S i x = 1, e n t o n c e s

l n (2) = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k + 1} = Σ u_{n}

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